$q.all returns 状态而不是值 - 为什么

Question

代码如下：

$q.all($q.when(3), $q.when(5)).then(function (values) {
    console.log(values);
});

输出如下：

{"$$state":{"status":1,"value":3}}

手册指出：

Returns a single promise that will be resolved with an array/hash of values, each value corresponding to the promise at the same index/key in the promises array/hash. If any of the promises is resolved with a rejection, this resulting promise will be rejected with the same rejection value.

所以我很困惑为什么没有返回值。

Answer 1

$q.all accepts an array or an object，所以如果你把它改成这样它应该可以工作：

$q.all([$q.when(3), $q.when(5)]).then(function (values) {
    console.log(values);
});